A radioactive nucleus n2 has 3 times the decay constant as compared to the decay constant of another radioactive nucleus n1. If initial number of both nuclei are the same, what is the ratio of number of nuclei of n2 to the number of nuclei of n1, after one half-life of n1?

N2=N0e−3λtN1=N0e−λtN2N1=e−2λtthalf life of N1: t=ln⁡2λN1N02=N0e−λtλt=ln⁡2t=ln⁡2λN2N1=e−2λln⁡2λN2N1=14

A radioactive nucleus n2 has 3 times the decay constant as compared to the decay constant of another radioactive nucleus n1. If initial number of both nuclei are the same, what is the ratio of number of nuclei of n2 to the number of nuclei of n1, after one half-life of n1?

N2=N0e−3λtN1=N0e−λtN2N1=e−2λtthalf life of N1: t=ln⁡2λN1N02=N0e−λtλt=ln⁡2t=ln⁡2λN2N1=e−2λln⁡2λN2N1=14

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A radioactive nucleus n₂ has 3 times the decay constant as compared to the decay constant of another radioactive nucleus n₁. If initial number of both nuclei are the same, what is the ratio of number of nuclei of n₂ to the number of nuclei of n₁, after one half-life of n₁?

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Detailed Solution

$\begin{array}{l} N_{2} = N_{0} e^{- 3 λt} \\ N_{1} = N_{0} e^{- λt} \\ \frac{N_{2}}{N_{1}} = e^{- 2 λt} \\ t_{{half life of N}_{1}} : t = \frac{\ln 2}{λ_{N_{1}}} \\ \frac{N_{0}}{2} = N_{0} e^{- λt} \\ λt = \ln 2 \\ t = \frac{\ln 2}{λ} \\ \frac{N_{2}}{N_{1}} = e^{- 2 λ \frac{\ln 2}{λ}} \\ \frac{N_{2}}{N_{1}} = \frac{1}{4} \end{array}$